Liouville property of fractional Lane-Emden equation in general unbounded domain

Theorem 1.1

Assume that α ∈ (0, 1), Ω ⊂ ℝ^N–1 × [0, +∞) is a C^0,1 domain verifying (𝓓) and Ω_∞ is given as (1.2) with 𝓞 = Ω. Then problem (1.1) has no nonnegative, nontrivial and bounded solutions if one of the following holds:

1. Ω_∞ = ℝ^N–1 and 0 < p < N−1+2αN−1−2α if N > 1 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α;

2. Ω_∞ ⊂ ℝ^N–1 is a star-shaped domain with respect to infinity and 1 ≤ p ≤ N−1+2αN−1−2α with N > 1 + 2α;

3. Ω_∞ = ℝ^N–2 × [0, +∞) and 1 < p < N−2+2αN−2−2α for N > 2 + 2α, otherwise for p > 1, 1 ≤ N ≤ 2 + 2α;

4. Ω_∞ is bounded, C² star-shape in ℝ^N–1 and N > 1 + 2α, p ≥ N−1+2αN−1−2α .

Remark 1.1

In the particular case that Ω is a cone such as {x = (x′, x_N) ∈ ℝ^N : x_N > θ|x′|} for some θ > 0, problem (1.1) has no nonnegative, nontrivial and bounded solutions.

Corollary 1.1

Under the assumptions of Theorem 1.1, we assume that the domain Ω_∞ ⊂ ℝ^N–2 × [0, +∞) is a C² domain verifying (𝓓) and Ω_∞,∞ is given by (1.2) with 𝓞_∞ = Ω_∞,∞. Then problem (1.1) does not admit nonnegative, nontrivial and bounded solutions if one of the following holds:

1. Ω_∞,∞ = ℝ^N–2 and 0 < p < N−2+2αN−2−2α if N > 2 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α;

2. Ω_∞,∞ ⊂ ℝ^N–2 is a star-shaped domain with respect to infinity and 1 ≤ p ≤ N−2+2αN−2−2α with N > 2 + 2α;

3. Ω_∞,∞ = ℝ^N–3 × [0, +∞) and 1 < p < N−3+2αN−3−2α for N > 3 + 2α, otherwise for p > 1, 1 ≤ N ≤ 3 + 2α;

4. Ω_∞,∞ is bounded, star-shape C² domain in ℝ^N–2 and N > 2 + 2α, p ≥ N−2+2αN−2−2α .

Proposition 2.1

Under the hypotheses of Theorem 1.1, let u be a nonnegative nontrivial solution of (1.1), then

Proof

We divide the proof into two steps.

1. By the assumption of Ω ⊂ R+N , we may assume

   λ0=inf{λ≥0:wλ≥0inΣλ}.

   The purpose of this step is to show that if λ > λ₀ is close to λ₀, then w_λ > 0 in Σ_λ. To this end, let Σλ− = {x ∈ Σ_λ | w_λ(x) < 0}, we first claim that if λ > λ₀ is close to λ₀, then

   Σλ−=∅. (2.5)

   By contradiction, we assume (2.5) is not true, that is Σλ− ≠ ∅. We denote

   wλ+(x)=wλ(x),x∈Σλ−,wλ+(x)=0,x∈RN∖Σλ−, (2.6)

   wλ−(x)=0,x∈Σλ−,wλ−(x)=wλ(x),x∈RN∖Σλ− (2.7)

   It is obvious that wλ+(x)=wλ(x)−wλ−(x) for all x ∈ ℝ^N. By direct computation, for x ∈ Σλ− , we have

   (−Δ)αwλ−(x)  = −∫(Ω∖Ωλ)∪(Ωλ∖Ω)wλ(z)|x−z|N+2αdz−∫(Σλ∖Σλ−)∪(Σλ∖Σλ−)λwλ(z)|x−z|N+2αdz−∫(Σλ−)λwλ(z)|x−z|N+2αdz=−I1−I2−I3.

   We look at each of these integrals separately. Since u = 0 in Ω_λ ∖ Ω and u_λ = 0 in Ω ∖ Ω_λ, then

   I1  = ∫Ωλ∖Ωuλ(z)|x−z|N+2αdz−∫Ω∖Ωλu(z)|x−z|N+2αdz=∫Ωλ∖Ωuλ(z)(1|x−z|N+2α−1|x−zλ|N+2α)dz≥0,

   by the fact that u_λ ≥ 0 and |x – z_λ| > |x – z| for all x ∈ Σλ− and z ∈ Ω_λ ∖ Ω.

   In order to fix the sign of I₂, note that w_λ(z_λ) = –w_λ(z) for any z ∈ ℝ^N and then

   I2  = ∫Σλ∖Σλ−wλ(z)|x−z|N+2αdz+∫Σλ∖Σλ−wλ(zλ)|x−zλ|N+2αdz=∫Σλ∖Σλ−wλ(z)(1|x−z|N+2α−1|x−zλ|N+2α)dz≥0,

   since w_λ ≥ 0 in Σ_λ ∖ Σλ− and |x – z_λ| > |x – z| for all x ∈ Σλ− and z ∈ Σ_λ ∖ Σλ− . Finally, since w_λ(z) < 0 for z ∈ Σλ− , we have

   I3=∫Σλ−wλ(zλ)|x−zλ|N+2αdz=−∫Σλ−wλ(z)|x−zλ|N+2αdz≥0.

   Hence, we obtain that for all λ > λ₀,

   (−Δ)αwλ−(x)≤0,∀x∈Σλ− (2.8)

   and then for x ∈ Σλ− ,

   (−Δ)αwλ+(x)≥(−Δ)αwλ(x)=(−Δ)αuλ(x)−(−Δ)αu(x). (2.9)

   Combining (1.1) with (2.9) and (2.6), we have that

   (−Δ)αwλ+(x)≥(−Δ)αuλ(x)−(−Δ)αu(x)=uλp(x)−up(x)=−φ(x)wλ+(x),x∈Σλ−,

   where

   φ(x)=−uλp(x)−up(x)uλ(x)−u(x),|φ|≤2p[uλp−1+up−1]≤2p∥u∥L∞p−1.

   Hence, we have

   Δαwλ+(x)≤φ(x)wλ+(x),x∈Σλ− (2.10)

   and observe that wλ+ = 0 in (Σλ−)c, then we have that

   supx∈Σλ−wλ+(x)≤cR(Σλ−)2α∥φwλ+∥L∞(Σλ−)≤cR(Σλ−)2α∥φ∥L∞(Σλ−)supx∈Σλ−wλ+(x),

   that is,

   1≤c2pR(Σλ−)2α∥u∥L∞p−1, (2.11)

   where c is a positive constant independent of Σλ− and

   R(Σλ−)=infr>0:|Br(x)∖Σλ−|≥12|Br(x)|,∀x∈Σλ−.

   Choosing λ > λ₀ close enough to λ₀, R( Σλ− ) is small, a contraction is derived from (2.11) and then

   wλ=wλ+≥0inΣλ−.

   But this is a contradiction with the definition of Σλ− , so we have that w_λ ≥ 0 in Σ_λ.

   We next claim that if w_λ ≥ 0 and w_λ ≢ 0 in Σ_λ, then w_λ > 0 in Σ_λ. Assuming this claim is true, we complete the proof, since the function u is positive in Ω and u = 0 on ∂Ω, so that w_λ is positive on ∂Ω ∩ ∂Σ_λ and then, by continuity w_λ ≢ 0 in Σ_λ.

   Now we prove above claim. In fact, assume there exists x₀ ∈ Σ_λ such that w_λ(x₀) = 0, that is, u_λ(x₀) = u(x₀). One hand we have that

   (−Δ)αwλ(x0)=(−Δ)αuλ(x0)−(−Δ)αu(x0)=uλp(x0)−up(x0)=0. (2.12)

   On the other hand, let A_λ = {(x₁, x′) ∈ ℝ^N | x₁ > λ}, since w_λ(z_λ) = –w_λ(z) for any z ∈ ℝ^N and w_λ(x₀) = 0, we find

   (−Δ)αwλ(x0)  =  −∫Aλwλ(z)|x0−z|N+2αdz−∫Aλwλ(zλ)|x0−zλ|N+2αdz=−∫Aλwλ(z)(1|x0−z|N+2α−1|x0−zλ|N+2α)dz.

   Since |x₀ – z_λ| > |x₀ – z| for z ∈ A_λ, w_λ(z) ≥ 0 and w_λ(z) ≢ 0 in A_λ, then

   (−Δ)αwλ(x0)<0, (2.13)

   which contradicts (2.12).

2. Our purpose of this step is to move the planes forward for any λ > λ₀ up to

   λ1:=sup{λ>λ0|wμ>0inΣμforanyμ∈(λ0,λ)}=+∞.

   By contradiction, we assume that λ₁ < +∞. Since for any λ ∈ (0, λ₁), we have that w_λ > 0 in Σ_λ, by the continuity, we derive that w_λ1 ≥ 0 in Σ_λ1 and w_λ1 ≢ 0 in Σ_λ1. Thus, by the claim just proved above, we have w_λ1 > 0 in Σ_λ1.

   Now we claim that if w_λ > 0 in Σ_λ1, then there exists ϵ > 0 such that w_λ > 0 in Σ_λ for all λ ∈ [λ₁, λ₁ + ϵ). This claim directly implies that λ₁ = +∞, completing Step 2.

   In fact, if λ₁ < +∞, then under our hypotheses, we have that Σ_λ is compact. Denote

   Dμ={x∈Σλ|dist(x,∂Σλ)≥μ}

   for μ > 0 small. Since w_λ1 > 0 in Σ_λ1 and D_μ is compact, then there exists μ₀ > 0 such that w_λ1 ≥ μ₀ in D_μ. By continuity of w_λ(x), for ϵ > 0 small enough and any λ ∈ (λ₁, λ₁ + ϵ), we have that

   wλ≥0inDμ.

   As a consequence,

   Σλϵ−⊂Σλϵ∖Dμ

   and R( Σλϵ− ) is small if ϵ and μ are small. Using (2.8) and proceeding as in Step 1, we have that for all x ∈ Σλϵ− ,

   (−Δ)αwλϵ+(x)  =  (−Δ)αuλϵ(x)−(−Δ)αu(x)−(−Δ)αwλϵ−(x)≥(−Δ)αuλϵ(x)−(−Δ)αu(x)=uλϵp(x)−up(x)=−φ(x)wλϵ+(x),

   where φ(x)=−uλϵp(x)−up(x)uλϵ(x)−u(x) is bounded. Since wλϵ+ = 0 in (Σλϵ−)c and | Σλϵ− | is small, for ϵ and μ small, from the analysis Step 1, we obtain that w_λϵ ≥ 0 in Σ_λϵ. Since λ_ϵ > 0 and w_λϵ ≢ 0 in Σ_λϵ, as before we have w_λϵ > 0 in Σ_λϵ, completing the proof of the claim. The proof ends.□

Proof of Theorem 1.1

We prove this argument by contradiction. Assume that u ≥ 0 is nontrivial solution of (1.1). When Ω verifies (𝓓), then by Proposition 2.1, we have that u satisfies (2.4). Let

then {u_m}_m is an increasing and bounded sequence of functions and satisfies

Note that Ω_t ⊂ Ω_t′ for 0 ≤ t ≤ t′ < +∞,

from the regularity results in [9, Theorem 2.1] and the stability property [9, Theorem 2.4], we obtain that

is a bounded classical solution to

By the increasing property of {u_m}_m, we have that

which implies that u_∞ is x_N-independent. Letting v_∞(x′) = u_∞(x′, x_N), by the standard argument, we have that

then v_∞ is a positive, bounded and classical solution for

subject to v_∞ = 0 in ℝ^N–1 ∖ Ω_∞ if Ω_∞ ≠ ℝ^N–1.

A contradiction is derived by the fact that problem (2.14) has no positive bounded classical solution when

1. Ω_∞ = ℝ^N–1 and 0 < p < N−1+2αN−1−2α if N > 1 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α by [13, Theorem 1.2];

2. Ω_∞ ⊂ ℝ^N–1 is a star-shaped domain with respect to infinity and 1 ≤ p ≤ N−1+2αN−1−2α with N > 1 + 2α by [14, Theorem 1.5];

3. Ω_∞ = ℝ^N–2 × [0, +∞) and 1 < p < N−2+2αN−2−2α for N > 2 + 2α, otherwise for p > 1, 1 ≤ N ≤ 2 + 2α by [14, Theorem 1.2];

4. Ω_∞ is bounded, C² and star-shape in ℝ^N–1, N > 1 + 2α, and p ≥ N−1+2αN−1−2α . by [14, Corollary 1.2].

As a consequence, we obtain the nonexistence results of (1.1) in Theorem 1.1.□

Proof of Corollary 1.1

When problem (1.1) has a positive solution u, then Proposition 2.1 and regularity results guarantee that v_∞(x′) = lim_m→+∞u(x′, x_N + m) is a positive, bounded and classical solution for problem

Furthermore, we note that v_∞,∞(x″) = lim_m→+∞ v_∞(x″, x_N–1 + m) is a positive, bounded and classical solution of

subject to v_∞,∞ = 0 in ℝ^N–2 ∖ Ω_∞,∞ if Ω_∞,∞ ≠ℝ^N–2. Then the same conclusion is obtained as the proof of Theorem 1.1.□